khò

quá khó

Mk biết nhưng mk ngại  viết lắm. Sorry nha

x + y = x . y

⇒ x + y − x . y = 0

⇒ x ( 1 − y ) + y = 0 

⇒ x ( 1 − y ) + ( y − 1 ) = −1 

⇒  ( 1 − y ) ( x − 1 ) = −1 

Ta có bảng sau : 

Vậy (x;y) thuộc (0;0);(2;2)

\(x+y=xy\)\(\Leftrightarrow xy-x-y=0\)

\(\Leftrightarrow x\left(y-1\right)-\left(y-1\right)=1\)( cộng 2 vế với 1 )

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)=1\)

Lập bảng giá trị ta có: 

Vậy \(x=y=0\)hoặc \(x=y=2\)

\(1,x:\left(-\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{3}\right)\\ \Leftrightarrow x=\left(-\dfrac{1}{3}\right)\times\left(-\dfrac{1}{3}\right)^3\\ \Leftrightarrow x=\left(-\dfrac{1}{3}\right)^4=\dfrac{1}{81}\\ 2,\left(\dfrac{4}{5}\right)^5.x=\left(\dfrac{4}{5}\right)^7\\ \Leftrightarrow x=\left(\dfrac{4}{5}\right)^7:\left(\dfrac{4}{5}\right)^5=\left(\dfrac{4}{5}\right)^{7-5}=\left(\dfrac{4}{5}\right)^2=\dfrac{16}{25}\)

\(3,\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(4,\left(3x+1\right)^3=-27\\ \Leftrightarrow\left(3x+1\right)^3=\left(-3\right)^3\\ \Leftrightarrow3x+1=-3\\ \Leftrightarrow3x=-4\\ \Leftrightarrow x=-\dfrac{4}{3}\)

\(5,\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\\ \Leftrightarrow x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\\ \Leftrightarrow x=\left(\dfrac{1}{2}\right)^{5-2}=\left(\dfrac{1}{2}\right)^3=\dfrac{1}{8}\)

\(6,\left(-\dfrac{1}{3}\right)^3.x=\dfrac{1}{81}\\ \Leftrightarrow\left(-\dfrac{1}{3}\right)^3.x=\left(-\dfrac{1}{3}\right)^4\\ \Leftrightarrow x=\left(-\dfrac{1}{3}\right)^4:\left(-\dfrac{1}{3}\right)^3=-\dfrac{1}{3}\)

\(7,\left(2x-3\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(8,\left(x-\dfrac{2}{3}\right)^3=\dfrac{1}{27}\\ \Leftrightarrow\left(x-\dfrac{2}{3}\right)^3=\left(\dfrac{1}{3}\right)^3\\ \Leftrightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)

`@` `\text {Ans}`

`\downarrow`

(Vế 1)

`1.`

`x \div(-1/3)^3 =-1/3`

`=> x= (-1/3) \times (-1/3)^3`

`=> x= (-1/3)^4`

`2.`

`(4/5)^5 *x = (4/5)^7`

`=> x = (4/5)^7 \div (4/5)^5`

`=> x=(4/5)^2`

`3.`

`(x+1/2)^2 =1/16`

`=> (x+1/2)^2 = (+-1/4)^2`

`=>`\(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{1}{4}-\dfrac{1}{2}\\x=-\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

`4.`

`(3x+1)^3 = -27`

`=> (3x+1)^3 = (-3)^3`

`=> 3x+1=-3`

`=> 3x=-3-1`

`=> 3x =-4`

`=> x=-4/3`

`5.`

`(1/2)^2*x=(1/2)^5`

`=> x=(1/2)^5 \div (1/2)^2`

`=> x=(1/2)^3`

`6.`

`(-1/3)^3*x=1/81`

`=> (-1/3)^3*x = (1/3)^4`

`=> x= (1/3)^4 \div (-1/3)^3`

`=> x=(-1/3)`

`7.`

`(2x-3)^2 = 16`

`=> (2x-3)^2 = (+-4)^2`

`=>`\(\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=7\\2x=-1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

`8.`

`(x-2/3)^3 = 1/27`

`=> (x-2/3)^3 = (1/3)^3`

`=> x-2/3=1/3`

`=> x=1/3 + 2/3`

`=> x=1`

Tấm vải đỏ dài:

84:(3+4)x3 =36 m

Tấm vải xanh dài:

84-36=48m

Đáp số : Tấm vải đỏ:36m

Tấm vải xanh:48m

Học tốt trong mùa dịch nha!

Trl :

Bạn kia làm đúng rồi nhé !

Học tốt nhé bạn @

Ta có:

12=1.12=2.6=3.4=4.3=6.2.12.1

và: 2x-1 là Ư lẻ của 12

=> 2x-1 E {1;3}

+) 2x-1=1=>2x=1+1=2

=>x=1

=>y+3=12=>y=9

Vậy x=1;y=9

+) 2x-1=3=>2x=3+1=4=>x=4:2=2

=> y+3=12:3=4

=>y=1

Vậy y=1;x=2

Câu 1 đường link câu này mk lm tương tư nhé

https://olm.vn/hoi-dap/detail/155610978.html

Băng Băng 2k6 giúp vs

1 + 1 = 2 nha bạn . 

1 + 1 = 2 nha bạn .